AMC12 2006 A

AMC12 2006 A · Q17

AMC12 2006 A · Q17. It mainly tests Quadratic equations, Circle theorems.

Square $ABCD$ has side length $s$, a circle centered at $E$ has radius $r$, and $r$ and $s$ are both rational. The circle passes through $D$, and $D$ lies on $\overline{BE}$. Point $F$ lies on the circle, on the same side of $\overline{BE}$ as $A$. Segment $AF$ is tangent to the circle, and $AF=\sqrt{9+5\sqrt{2}}$. What is $r/s$?

正方形 $ABCD$ 边长为 $s$，以 $E$ 为中心的圆半径为 $r$，且 $r$ 和 $s$ 均为有理数。该圆经过 $D$，且 $D$ 位于 $\overline{BE}$ 上。点 $F$ 位于圆上，位于 $\overline{BE}$ 与 $A$ 同侧。线段 $AF$ 切于圆，且 $AF=\sqrt{9+5\sqrt{2}}$。求 $r/s$？

(A) \frac{1}{2} \frac{1}{2}

(B) \frac{5}{9} \frac{5}{9}

(D) \frac{5}{3} \frac{5}{3}

(E) \frac{9}{5} \frac{9}{5}

Answer

Correct choice: (B)

正确答案：（B）

Solution

One possibility is to use the coordinate plane, setting $B$ at the origin. Point $A$ will be $(0,s)$ and $E$ will be $\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)$ since $B, D$, and $E$ are collinear and contain a diagonal of $ABCD$. The Pythagorean theorem results in \[AF^2 + EF^2 = AE^2\] \[r^2 + \left(\sqrt{9 + 5\sqrt{2}}\right)^2 = \left(\left(s + \frac{r}{\sqrt{2}}\right) - 0\right)^2 + \left(\left(s + \frac{r}{\sqrt{2}}\right) - s\right)^2\] \[r^2 + 9 + 5\sqrt{2} = s^2 + rs\sqrt{2} + \frac{r^2}{2} + \frac{r^2}{2}\] \[9 + 5\sqrt{2} = s^2 + rs\sqrt{2}\] This implies that $rs = 5$ and $s^2 = 9$; dividing gives us $\frac{r}{s} = \frac{5}{9} \Rightarrow B$.

一种方法是使用坐标平面，令 $B$ 在原点。点 $A$ 为 $(0,s)$，并且由于 $B, D, E$ 共线且该直线包含 $ABCD$ 的一条对角线，所以 $E$ 为 $\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)$。由勾股定理得到 \[AF^2 + EF^2 = AE^2\] \[r^2 + \left(\sqrt{9 + 5\sqrt{2}}\right)^2 = \left(\left(s + \frac{r}{\sqrt{2}}\right) - 0\right)^2 + \left(\left(s + \frac{r}{\sqrt{2}}\right) - s\right)^2\] \[r^2 + 9 + 5\sqrt{2} = s^2 + rs\sqrt{2} + \frac{r^2}{2} + \frac{r^2}{2}\] \[9 + 5\sqrt{2} = s^2 + rs\sqrt{2}\] 这意味着 $rs = 5$ 且 $s^2 = 9$；相除得到 $\frac{r}{s} = \frac{5}{9} \Rightarrow B$。

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